Question

The mirror image of the curve $\arg \left(\frac{z-3}{z-i}\right)=\frac{\pi }{6}$ in the real axis is

• A. $\arg \left(\frac{z+3}{z+i}\right)=\frac{\pi }{6}$
• B. $\arg \left(\frac{z-3}{z+i}\right)=\frac{\pi }{6}$
• C. $\arg \left(\frac{z+i}{z+3}\right)=\frac{\pi }{6}$
• D. $\arg \left(\frac{z+i}{z-3}\right)=\frac{\pi }{6}$

Answer 1

Answer: (D)

The image of $z$ in the real axis is $\bar{z}$
$\therefore$ The image of given curve is given by $\arg \left(\frac{\bar{z}-3}{\bar{z}-i}\right)=\frac{\pi }{6}$
But $\arg \bar{z}=-\arg z$
$\therefore \arg \left(\frac{\bar{z}-3}{\bar{z}-i}\right)=\frac{\pi }{6}$
$\Rightarrow \arg \left(\frac{z-3}{z+i}\right)=-\frac{\pi }{6}$
$\Rightarrow \arg \left(\frac{z+i}{z-3}\right)=\frac{\pi }{6}$